A051707

[kohmoto]

    Hello, Seqfans.
    I explain the origin of A051707.

    Cristian wrote :
>
>> I rewrite the condition as follows.
>
>> Pair (i,j) must satisfy 0<i, 0<j, and if i=1 then j=1.
>> The second condition is equivalence as that if (i,j) is not
>> (1,1) then i is not 1.
>
> Implying that (1,1) is allowed as a factor.
>
> That cannot be the case, otherwise there would be infinitely many
> factorizations:
>
> (2,2)
> (2,2)*(1,1)
> (2,2)*(1,1)*(1,1)
> ...
>
> So a better restriction is that: 1<i, 0<j
>

    Ordinary, they don't count factorizations which have the unit.
    So, my definition and yours are equivalent.

    If we consider only about the sequence, I agree with your opinion.

    But the "Algebra" is an abstraction of the calculation of (m,n) Amicable 
Number.
    In a system of the calculation, the "unit" exists.
    So, I like my definition.

    [Definition of (m,n) Amicable Number]

         Sigma(x_i) = m/n*( Sum_{1<=i<=n} x_i ) , i=1 to n      -D-
         n-ple (x_i) which satisfies D is called (m,n) Amicable Number.

    Example.1.: Amicable Pair = (2,2) AN . AN is for Amicable Number.
         Sigma(x_i) = 2/2*( Sum_{1<=i<=2} x_i ) = x_1+x_2 , i=1 and 2

    Example.2. 1/2 Amicable Quadruple = (2,4) AN
         Sigma(x_1) = Sigma(x_2) = Sigma(x_3) = Sigma(x_4) = 
1/2*(x_1+x_2+x_3+x_4)

    Example.3. Perfect Number = (2,1) AN
         Sigma(x_1)=2*x_1

    [Factorization of (m,n) AN]
    If
         (m,n) AN : Sigma(x_i) = m/n*( Sum_{1<=i<=n} x_i ) , i=1 to n
         (r,s) AN : Sigma(x_j) = r/s*( Sum_{1<=j<=n} x_j ) , j=1 to s
         GCD(x_i,x_j)=1 for all i,j     Then
                         Sigma(x_i*x_j)=(m*r)/(n*s)*( Sum_{1<=i<=n , 
1<=j<=n} x_i*x_j ) , i=1 to n , j=1 to s .
    It becomes the definition of (m*r,n*s) AN.
    So, it is possible to define as follows.
         (m,n) AN*(r,s) AN  =   (m*r,n*s) AN      n*s-ple 
(x_{i,j})=(x_i*x_j)

    Example : Amicable Quadruple = (4,4) AN = (2,2) AN*(2,2) AN = (2,1) AN* 
(2,4) AN
    See this page. Examples of these factorizations exist.
         http://mathworld.wolfram.com/AmicableQuadruple.html

     A051707 is the factorization of (m,n) AN. The case of m=n.
     (1,1) corresponds (1,1) AN which satisfies Sigma(x)=x ; x=1 .

> Using that I get for the first several terms:
>
> 1 1 1 3 1 5 1 8 3 5 1 23 1 5 5 23 1 23 1 23 5 5 1 91 3 5 8 23 ...
>
> This matches most of A051707
> ID Number: A051707
> URL:       http://www.research.att.com/projects/OEIS?Anum=A051707
> Sequence:  1,1,1,3,1,5,1,8,3,5,1,21,1,5,5,23,1,21,1,21
> Name:      Number of factorizations of (n,n) into pairs (k,l).
>
> but is different for 12, 18 and 20
>
> The 23 factorizations of 12 are as follows:
>
> (12,12)
> 6 of the form (6,x)*(2,y)
> 6 of the form (4,x)*(3,y)
> 10 of the form (3,x)*(2,y)*(2,z)
>
> I can send an update if Yasutoshi can confirm that those values are
> errors and not the result of yet another rule.
>
    I think you are correct.


>> I want to know the formula.
>> I wish some one tell me it.
>
> I can help if you're not too picky about what a formula is.
>
> What I did was apply the setprod transform I recently described to an
> array of numbers a(m,n) such that a(1,n) = 0 and a(m,n) = 1 for all
> m>=2, n>=1
>
> The output array gives the number of factorizations of (m,n):
>
> 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 2 2 2 3 2 3 2 3 3 3 2 4 2 3 3 4 2 4 2 4 3 3 2 5
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 2 3 3 4 3 5 3 5 4 5 3 7 3 5 5 6 3 7 3 7 5 5 3 9
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 3 4 4 6 4 7 4 8 6 7 4 11 4 7 7 10 4 11 4 11 7 7 4 15
> 2 2 2 3 2 3 2 3 3 3 2 4 2 3 3 4 2 4 2 4 3 3 2 5
> 2 3 3 4 3 5 3 5 4 5 3 7 3 5 5 6 3 7 3 7 5 5 3 9
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 4 7 7 11 7 14 7 15 11 14 7 23 7 14 14 20 7 23 7 23 14 14 7 33
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> ...
>
    Yes, it should be an array.


> Then I take the diagonal b(n,n) to produce the sequence.
>
> The mathematics of this is to treat the input array as a
> 2 dimensional Dirichlet generating function:
>
> A(s,t) = SUM {m,n=1 to infinity} a(m,n)/(m^s*n^t)
>
> apply setprod as follows
>
> B(s,t) = exp(A(s,t)/1 + A(2*s,2*t)/2 + A(3*s,3*t)/3 + ...)
>
>       = PROD {m,n=1 to infinity} (1/(1-a(m,n)/(m^s*n^t)))
>
> If you want to know the algorithm I actually use or the code I
> use to execute it, you should probably contact me off line unless
> there are several on the list that are interested.
>
    You know that I am an amateur mathematician, so I have few knowledge 
about "2 dimensional Dirichlet generating function".
    I want to learn about it.

    Yasutsohi